SOLUTION: Solve each equation. If a solution is extraneous, so indicate. (5)/(2z+z-3)-(2)/(2z+3)=(z+1)/(z-1)-(1)

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Solve each equation. If a solution is extraneous, so indicate. (5)/(2z+z-3)-(2)/(2z+3)=(z+1)/(z-1)-(1)      Log On


   

Question 59829This question is from textbook Intermediate Algebra
: Solve each equation. If a solution is extraneous, so indicate.
(5)/(2z+z-3)-(2)/(2z+3)=(z+1)/(z-1)-(1) This question is from textbook Intermediate Algebra

Answer by funmath(2933) About Me  (Show Source):
You can put this solution on YOUR website!
Solve each equation. If a solution is extraneous, so indicate.
(5)/(2z+z-3)***-(2)/(2z+3)=(z+1)/(z-1)-(1)
***I think you meant to type (2z^2+z-3)
5%2F%282z%5E2%2Bz-3%29-2%2F%282z%2B3%29=%28z%2B1%29%2F%28z-1%29-1 Factor the denominators.
5%2F%28%282z%2B3%29%28z-1%29%29-2%2F%282z%2B3%29=%28z%2B1%29%2F%28z-1%29-1
The LCD is(2z+3)(z-1), if you get a restriced value as an answer, it's extraneous.
2z+3=0--->2z=-3--->z=-3/2 Is a restricted value that will give you a 0 in the denominator.
z-1=0-->z=1 Is also a restricted value.
Multiply everything by the LCD:


5-2%28z-1%29=%28z%2B1%29%282z%2B3%29-1%282z%2B3%29%28z-1%29
5-2z%2B2=2z%5E2%2B3z%2B2z%2B3-2z%5E2-z%2B3
%285%2B2%29-2z=%282-2%29z%5E2%2B%283%2B2-1%29z%2B%283%2B3%29
7-2z=4z%2B6
7-6-2z%2B2z=4z%2B2z%2B6-6
1=6z
1%2F6=z
That's not a restrted value so the solutions is:highlight%28z=1%2F6%29
Happy Calculating!!!